log \(\left[\cfrac{x+y}{3}\right]\) = \(\cfrac{1}2\) (log x + log y), then the value of \(\cfrac{x}y\) + \(\cfrac{y}x\) is ..........

A) 9 

B) 5

C) 7 

D) 11

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2 Answers

Correct option is (C) 7

Given that \(log\left(\frac{x+y}{3}\right)=\frac{1}{2} (log\,x+log\,y)\)

\(\Rightarrow\) \(2\,log\,(\frac{x+y}3)\) = log x + log y

\(\Rightarrow\) \(log\,(\frac{x+y}3)^2\) = log xy   \((\because\) log A + log B = log AB and n log a \(=log\,a^n)\)

\(\Rightarrow\) \((\frac{x+y}3)^2\) = xy   (By comparing)

\(\Rightarrow\) \(x^2+y^2+2xy\) = 9xy

\(\Rightarrow\) \(\frac{x^2}{xy}+\frac{y^2}{xy}+\frac{2xy}{xy}=\frac{9xy}{xy}\)   (Dividing both sides by xy)

\(\Rightarrow\) \(\frac xy+\frac yx+2=9\)

\(\Rightarrow\) \(\frac{x}{y}+\frac{y}{x}\) = 9 - 2 = 7

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Correct option is C) 7

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